C

Step 1: Analyze projectile motion.
The shell fired from the gun behaves like a projectile. When fired with a muzzle velocity $u$, the horizontal distance $R$ travelled can be expressed as a function of time and initial velocity.
Step 2: Determine time of flight to height h.
The vertical motion of the shell follows the equation: $$ h = u_y t - \frac{1}{2} g t^2 $$ where $u_y = u \sin \theta$ (the vertical component of the velocity), $g$ is the acceleration due to gravity, and $t$ is the time of flight. In this case, we need to find the time taken to reach height $h$. Rearranging gives us: $$ t = \frac{u_y + \sqrt{u_y^2 + 2gh}}{g} $$
Step 3: Horizontal motion.
The horizontal distance travelled by the shell is: $$ R = u_x t $$ where $u_x = u \cos \theta$ (the horizontal component of the velocity).
Substituting for $t$ gives us maximum range as a function of $ heta$. To maximize $R$, we can derive it based on the angle of projection. The optimal angle for maximizing range in projectile motion is $45^{\circ}$.
Step 4: Final equations.
Thus, at this angle: $$ R = \frac{u^2 \sin(2\theta)}{g} $$
Step 5: Conclusion.
After considering the height $h$ above which the plane flies, the maximum horizontal distance changes based on the initial height $h$ of the target. The formula can effectively represent this scenario relating to gunfire and aerial targets. Thus, the answer is found to be related to: $$ R = \frac{u^2}{g} + \sqrt{2h}$$. Therefore, the correct answer is option C.